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letanque
43d

Why do all these quantum computers I see look like steampunk chandeliers?

And what's the real value going to be?

Comments
  • 11
    Calculations you couldn't even imagine. And as for looks, that's probably due to the extreme temperatures at which they operate
  • 8
    Cooling, mostly. Everything about quantum computer design is dominated by the effort of cooling the system and keeping it stable.

    Advantages? Quantum algorithms which shred classical ones for fairly important problems. A quantum algorithm has a much richer kind of information and transformation steps, so it can go beyond classical limits (easiest simplified explanation I can think of). These aren't going to be what you use to play the newest games, but more like another class of specialized accelerators. In the near future anyway, who knows about distant future.

    For a very simple example of what this "richness" of information is like fundamentally, consider the unit of classical information, the bit. You know that it's either 0, or 1. Two possibilities. That's all a bit can represent.

    The quantum unit is the qubit which you can think of as being a "mixture" of "0-ness" and "1-ness" (it's *not* a state "in between" 0 and 1 classically, so it's not some wonky value like 0.57, kinda confusing but eh). To "see" the value of a qubit you need to measure it, and when you do it collapses to either 0 or 1. You can do operations on it without measuring it though, and that's where it becomes powerful.

    Technically, the 0-ness corresponds to the probability that the qubit outputs 0 when measured, and the 1-ness corresponds to the probability that the qubit outputs 1 when measured. So a classical bit could be a qubit with no 0-ness (so it'll always be 1 when measured) or no 1-ness (so it'll always be 0 when measured), but it can also mix, so you get states in which for example in half the measurements you get 0 and other half you get 1 (assuming you repeat the same measurement on identical qubits), or 2/3 of them measure to 0 and 1/3 measure to 1 or whatever.

    These mixed states (along with similarly "richer" algorithms and transformations) allow qubit-based systems to be far more capable than bit-based systems at a fundamental level. There's just more to them.
  • 2
    Basically, quantum computers are a special type of supercomputers for very specific tasks and algorithms.
  • 2
    To add on the previous comments: the idea is to overlay operations to not do them sequentially, but all at the same time.

    One very important application: if you have two very large prime numbers, it's easy to multiply them.

    But if you only have the product, not the factors, it's difficult to find the factors. You'd have to try a lot of numbers, and that would take an eternity even on super computers.

    That's what they call a one-way-function: easy to compute in forward direction, hard to do backwards. That's the thing at the heart e.g. of public key encryption such as PGP.

    However, if you could try all the divider numbers at the same time instead of sequentially, you'd get the factors back in no time. PGP would be broken!

    Whoever can first get a quantumcomputer ready for decryption will have an immense advantage over everyone else world-wide.
  • 8
    Better/different explanation, sort of.

    Imagine the universe as a simulation.

    Quantum mechanics utilizes a flaw/feature of the simulation where it offloads a little bit of processing to the host. As the host performs this, our simulation sees the result instantly. Quantum computers exploit this quirk to dramatically increase the amount of processing offloaded, and can give the host incredibly complex tasks to solve, which it does instantly. Setting these up is extremely difficult, so they’re only better for tasks that classical computers are terrible at — like finding enormously large factors. This task in particular they particularly excel at.

    There is obviously a lot more to it, but this should give you an idea!

    (Also, if the universe really is a simulation or not is up for grabs. I personally believe it is more than 50% likely for reasons I won’t get into here.)
  • 1
    Are we collapsing wave functions? Dibs on the eigenstate.
  • 4
    I think others already provided awesome and on-point answers, especially @RememberMe's explanation ticks all the checkboxes you need to know in a simplified form.

    I'll just add that since Quantum computers are essentially a massively parallel machines, it also means they are not the right solution to some problems. Like video games, because there things happening in sequence is a must.

    This is why you shouldn't expect them to be the, so to say: "next" thing, just another tool... For some time they will definitely just be used as accelerators or specialized machines.

    And yeah, all the pipes are cooling, but it's not because it gets too hot, It's rather because it operates at sub-zero temperatures at all times.

    See, Quantum particles are wobly when they have energy... To realy be able to use them you need to be close to absolute zero, as close as possible at least.... And That's hella hard
  • 0
    Because copper is great
  • 1
    @RememberMe interesting, I always thought that systems using a qubit just had 3 states. But exact states. Where a binary bit can be (0 OR 1) a quantum bit can be (0 OR 1 OR (0 AND 1)). Error correcting taking care of wonky maybe it's 1 maybe you got a improbable 0, hey stuff happens...
  • 1
    @hjk101 the mixed state sort of is like your 0 and 1, but there's a continuum of them, plus the whole measurement mechanism which is entirely unique to qubits (it's still a binary digit, so the measurement can only give a 0 or 1).

    For more detail, a qubit state is represented like
    a * |0> + b * |1>
    Where a and b are constants such that a^2 + b^2 = 1, and the weird 0 and 1 notation is just how we write the 0 and 1 state vectors. So a is 0-ness, b is 1-ness. a^2 (a squared) is the probability that you'd get a 0 when you measure a qubit, and b^2 is the probability that you'd get a 1 (why not just use the squared numbers directly in the representation? Because squaring loses information - think 2 and -2 having the same square, 4).

    These a and b constants store the information of the qubit. if a = 0, b = 1, we have the classical bit 1. If a = 1, b = 0, we have the classical bit 0. If a = b = 1/sqrt(2), (so the square is 1/2) we have the 50-50 case where it could measure out to be either. Quantum processes affect these a and b values directly via interference, wave function evolution in fields etc. and give rise to more-than-classical behavior.

    Also shows why simulation of qubit systems is so hard classically (at least naively). An interacting qubit system is represented as the probability of each combination, so for 2 qubits you have a number for 00, 01, 10, 11 each; for 3 qubits you need a number for all 8 combinations, for 4 it's 16, 5 it's 32, for n it's 2^n, so it's exponentially difficult to (naively) model quantum qubits with classical bits. And a and b have infinite precision realistically, but classical storage allows finite precision only, so that's much worse.
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